1.3 Sequences

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We can use the absolute value of a complex number to determine distance in the complex plane, and therefore to define what it means for a sequence of complex numbers to converge to a limit.

Definition (Distance)

The distance between complex numbers $z,w\in \mathbb{C}$ is $|z-w|$.

Figure 1: The distance between complex numbers $z$ and $w$ is defined to be $|z-w|$.

The distance between two complex numbers as we have defined it here is the same as the Euclidean distance between the corresponding vectors in ${\mathbb{R}}^2$.

Definition (Open Ball)

The open ball of radius $r>0$ centered at $z\in \mathbb{C}$ is the set $$\mathsf{B}(z,r)=\{w\in \mathbb{C}:|z-w|<r\}$$ of all complex numbers $w$ within a distance of $r$ from $z$.

Open balls give us a notion of nearness. If complex numbers belong to an open ball of small radius then they are all close to the center of the ball.

Figure 2: The open ball $\mathsf{B}(z,r)$ centered at $z$ of radius $r$ is the shaded region of the plane. The point $w$ belongs to the ball.

The crucial definition in this section is what it means for a sequence of complex numbers to converge to a limiting complex number. Recall that a sequence of complex numbers is just an indexed list $z_1,z_2,z_3,\dots$ of complex numbers. Formally, a sequence is any function from $\mathbb{N}$ to $\mathbb{C}$.

Definition (Convergence)

A sequence $z_n$ of complex numbers converges to a complex number $a$ if, for every $ϵ>0$ one can find $N\in \mathbb{N}$ such that $n\ge N$ implies $|z_n-a|<ϵ$.

When $z_n$ converges to $a$ we write ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{z}_n=a}$. This is equivalent to the real limit statement ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}|z_n-a|=0}$. We can interpret convergence in terms of open balls because $|z_n-a|<ϵ$ is equivalent to $z_n\in \mathsf{B}(a,ϵ)$.

Figure 3: A sequence $z_n$ converges to $a$ if and only if, for every $ϵ>0$ there is $N\in \mathbb{N}$ such that $z_n\in \mathsf{B}(a,ϵ)$ for all $n\ge N$.

Example

The sequence $z_n={\displaystyle \frac{1}{n+i{n}^2}}$ converges to 0.

Solution:

Fix $ϵ>0$. We must produce $N\in \mathbb{N}$ with the property that $|z_n-0|<ϵ$ whenever $n\ge N$. First we calclate $$\begin{array}{rl}|z_n-0|=\left|\frac{1}{n+i{n}^2}\right|& =\left|\frac{n-i{n}^2}{n^2+n^4}\right|\\ & =\sqrt{{\left(\frac{n}{n^2+n^4}\right)}^2+{\left(\frac{-n^2}{n^2+n^4}\right)}^2}\\ & =\sqrt{\frac{1}{n^2+n^4}}\\ & \le \frac{1}{n}\end{array}$$ so that, if $n\ge N>\frac{1}{ϵ}$ we automatically have $|z_n-0|<ϵ$.

Lemma

If ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{z}_n=a}$ and ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{w}_n=b}$ then the following all hold.

1. ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{z}_n+w_n=a+b}$
2. ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}c{z}_n=ca}$ for all $c\in \mathbb{C}$
3. ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{z}_n{w}_n=ab}$
4. ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}1/z_n=1/a}$ if $a\ne 0$

Proof:

The proofs of these results are exactly the same as in the real case. $◻$

Lemma

Let $z_n\in \mathbb{C}$ for all $n\in \mathbb{N}$ and write $z_n=x_n+i{y}_n$. Then $z_n$ converges if and only if $x_n$ and $y_n$ converge.

Proof:

Suppose that $z_n\to z$ and write $z=x+iy$. Then $$|x_n-x|\le \sqrt{|x_n-x{|}^2+|y_n-y{|}^2}=|z_n-z|\to 0$$ as $n\to \mathrm{\infty }$. Hence $x_n\to x$. A similar argument show that $y_n\to y$.

Conversely, suppose that $x_n$ converges to $a$ and $y_n$ converges to $b$. Then $$\underset{n\to \mathrm{\infty }}{lim}{z}_n=\underset{n\to \mathrm{\infty }}{lim}{x}_n+i\underset{n\to \mathrm{\infty }}{lim}{y}_n=a+ib$$ so that $z_n\to a+ib$. $◻$